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Bổ sung đề: C(6;2)
a: vecto AD=(xD+3;yD-6)
vecto BD=(xD-1;yD+2)
vecto CD=(xD-6;yD-2)
Theo đề, ta có: \(\left\{{}\begin{matrix}x_D+3+2\left(x_D-1\right)-4\left(x_D-6\right)=0\\y_D-6+2\left(y_D+2\right)-4\left(y_D-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_D+3+2x_D-2-4x_D+24=0\\y_D-6+2y_D+4-4y_D+8=0\end{matrix}\right.\)
=>D(25;6)
\(m\overrightarrow{a}=m\left(-1;-2\right)=\left(-m;-2m\right)\)
\(n\overrightarrow{b}=n\left(1;-3\right)=\left(n;-3n\right)\)
\(\Rightarrow m\overrightarrow{a}+n\overrightarrow{b}=\left(-m+n;-2m-3n\right)\)
\(\Rightarrow\left\{{}\begin{matrix}-m+n=2\\-2m-3n=-4\end{matrix}\right.\) \(\Rightarrow m-n=-2\) (đảo dấu pt đầu là ra, ko cần giải hẳn ra m; n)
Ta có: \(\left\{{}\begin{matrix}\overrightarrow{a}=m\overrightarrow{u}+\overrightarrow{v}=\left(4m+1;m+4\right)\\\overrightarrow{b}=\overrightarrow{i}+\overrightarrow{j}=\left(1;1\right)\end{matrix}\right.\)
Yêu cầu bài toán <=> cos\(\left(\overrightarrow{a};\overrightarrow{b}\right)\)=cos45o =\(\dfrac{\sqrt{2}}{2}\)
<=> \(\dfrac{\left(4m+1\right)+\left(m+4\right)}{\sqrt{2}\sqrt{\left(4m+1\right)^2+\left(m+4\right)^2}}=\dfrac{\sqrt{2}}{2}\)
<=> \(\dfrac{5\left(m+1\right)}{\sqrt{2}\sqrt{17m^2+16+17}}=\dfrac{\sqrt{2}}{2}\)
<=> \(5\left(m+1\right)=\sqrt{17m^2+16m+17}\) <=>\(\left\{{}\begin{matrix}m+1\ge0\\25m^2+50m+25=17m^2+16m+17\end{matrix}\right.\)
<=> m=\(-\dfrac{1}{4}\)
\(cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\dfrac{1\cdot\left(-1\right)+\left(-2\right)\cdot\left(-3\right)}{\sqrt{1^2+2^2}\cdot\sqrt{1^2+3^2}}=\dfrac{5}{\sqrt{5}\cdot\sqrt{10}}=\dfrac{5}{\sqrt{50}}=\dfrac{1}{\sqrt{2}}\)
Lời giải:
Gọi \(\overrightarrow{d}=(x,y)\). Theo bài ra ta có:
\(\left\{\begin{matrix} \overrightarrow{a}.\overrightarrow{d}=4\\ \overrightarrow{b}.\overrightarrow{d}=-2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -2x+3y=4\\ 4x+y=-2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{-5}{7}\\ y=\frac{6}{7}\end{matrix}\right.\)
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